question_answer

Look at the figure, O is the centre of in circle inscribed in\[\Delta \,PQR\].

Find \[\angle QOR\] if \[\angle QPR=50{}^\circ \]

A)  \[95{}^\circ \]                         

B)  \[125{}^\circ \]          

C)  \[115{}^\circ \]       

D)  \[125{}^\circ \]

Correct Answer: C

Solution :

(c): Make constructions as shown: By property of in circle, PO, QO and RO, and \[\angle \] bisectors \[\Rightarrow \] at point P, we have angles as \[25{}^\circ \] each; \[\angle OUP=90{}^\circ \] \[\therefore \] In \[\Delta \text{ }POU,\]\[\angle POU=90{}^\circ -25{}^\circ =65{}^\circ =y\] Now at point \[O,x+x+y+y+z+z=360{}^\circ \] \[\Rightarrow \]\[2\left( x+z \right)=360{}^\circ -2y-230{}^\circ \] \[\Rightarrow \]\[\angle QOR=x+z=115{}^\circ \]